How do you find the length of the curve $y = ln | sec x |$ $y=lnabs(secx)$ from $0 \leq x \leq \frac{π}{4}$ $0<=x<=pi/4$ ?

Question

4938 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-02-02T12:39:42

$ln (\sqrt{2} + 1) = 0.8814$ $ln(sqrt2+1)=0.8814$ length units, nearly..

In the given interval, $| sec x | = sec x$ $|secx|=secx$ .

$y'=1/secx (secx)'=secxtanx/secx=tanx$ 1

Length = $int sqrt(1+(y')^2) dx$ ,

with $y = ln|secx|$ and x from $0 to pi/4$

$=int sqrt(1+tan^2x) dx$ , for the limits

$=int secx dx$ , for the limita

$=[ln(secx+tanx)],$ between $x = 0 and pi/4$

$=ln(sec(pi/4)+tan(pi/4))=ln(sqrt2+1)$

How do you find the length of the curve y=lnabs(secx)y=ln|secx|y=lnabs(secx) from 0<=x<=pi/40≤x≤π40<=x<=pi/4?